Virudhunagar District
Common First Mid Term Test - 2024
Standard 10 - MATHS
PART I: Multiple Choice Questions
1) If \(n(A \times B) = 6\) and \(A = \{1, 3\}\) then \(n(B)\) is
- 1
- 2
- 3
- 6
We know that \(n(A \times B) = n(A) \times n(B)\).
Given, \(n(A \times B) = 6\) and \(A = \{1, 3\}\), so \(n(A) = 2\).
Substituting the values: \(6 = 2 \times n(B)\).
Therefore, \(n(B) = \frac{6}{2} = 3\).
The correct option is (c) 3.
2) \(f(x) = (x+1)^3 - (x-1)^3\) represents a function which is
- linear
- cubic
- reciprocal
- quadratic
We expand the expression for \(f(x)\):
Using the identity \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) and \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\).
\((x+1)^3 = x^3 + 3(x^2)(1) + 3(x)(1^2) + 1^3 = x^3 + 3x^2 + 3x + 1\)
\((x-1)^3 = x^3 - 3(x^2)(1) + 3(x)(1^2) - 1^3 = x^3 - 3x^2 + 3x - 1\)
\(f(x) = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1)\)
\(f(x) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1\)
\(f(x) = (x^3 - x^3) + (3x^2 + 3x^2) + (3x - 3x) + (1 + 1)\)
\(f(x) = 6x^2 + 2\)
Since the highest power of \(x\) is 2, the function is quadratic.
The correct option is (d) quadratic.
3) The sum of the exponents of the prime factors in the prime factorization of 1729 is
- 1
- 2
- 3
- 4
We find the prime factorization of 1729.
1729 is not divisible by 2, 3, 5. Let's try 7.
\(1729 \div 7 = 247\)
Now, let's factorize 247. It is not divisible by 7, 11. Let's try 13.
\(247 \div 13 = 19\)
19 is a prime number.
So, the prime factorization of 1729 is \(7^1 \times 13^1 \times 19^1\).
The exponents of the prime factors are 1, 1, and 1.
The sum of the exponents = \(1 + 1 + 1 = 3\).
The correct option is (c) 3.
4) If 6 times of 6th term of an A.P is equal to 7 times the 7th term, then the 13th term of the A.P is
- 0
- 6
- 7
- 13
Let the A.P. have first term 'a' and common difference 'd'.
The nth term is given by \(t_n = a + (n-1)d\).
Given, \(6 \times t_6 = 7 \times t_7\).
\(6(a + (6-1)d) = 7(a + (7-1)d)\)
\(6(a + 5d) = 7(a + 6d)\)
\(6a + 30d = 7a + 42d\)
\(6a - 7a = 42d - 30d\)
\(-a = 12d \implies a + 12d = 0\)
We need to find the 13th term, \(t_{13}\).
\(t_{13} = a + (13-1)d = a + 12d\)
From our previous result, we know \(a + 12d = 0\).
Therefore, \(t_{13} = 0\).
The correct option is (a) 0.
5) If \(f : A \to B\) is a bijective function and if \(n(B) = 7\), then \(n(A)\) is equal to
- 7
- 49
- 1
- 14
A bijective function is a function that is both one-to-one (injective) and onto (surjective).
For a function to be bijective between two finite sets A and B, the number of elements in both sets must be equal.
That is, \(n(A) = n(B)\).
Given \(n(B) = 7\), it follows that \(n(A) = 7\).
The correct option is (a) 7.
6) The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
- 2025
- 5220
- 5025
- 2520
We need to find the Least Common Multiple (LCM) of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
First, find the prime factorization of each number:
- \(1 = 1\)
- \(2 = 2\)
- \(3 = 3\)
- \(4 = 2^2\)
- \(5 = 5\)
- \(6 = 2 \times 3\)
- \(7 = 7\)
- \(8 = 2^3\)
- \(9 = 3^2\)
- \(10 = 2 \times 5\)
To find the LCM, we take the highest power of each prime factor present in the factorizations.
Highest power of 2 is \(2^3\).
Highest power of 3 is \(3^2\).
Highest power of 5 is \(5^1\).
Highest power of 7 is \(7^1\).
LCM = \(2^3 \times 3^2 \times 5^1 \times 7^1 = 8 \times 9 \times 5 \times 7 = 72 \times 35 = 2520\).
The correct option is (d) 2520.
7) The next term of the sequence \(\frac{1}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}, \dots\) is
- \(\frac{1}{24}\)
- \(\frac{1}{27}\)
- \(\frac{2}{3}\)
- \(\frac{1}{81}\)
Let's check the ratio between consecutive terms to see if it's a Geometric Progression (GP).
Ratio 1: \(\frac{t_2}{t_1} = \frac{1/8}{1/16} = \frac{1}{8} \times 16 = 2\)
Ratio 2: \(\frac{t_3}{t_2} = \frac{1/12}{1/8} = \frac{1}{12} \times 8 = \frac{8}{12} = \frac{2}{3}\)
Ratio 3: \(\frac{t_4}{t_3} = \frac{1/18}{1/12} = \frac{1}{18} \times 12 = \frac{12}{18} = \frac{2}{3}\)
It appears that from the second term onwards, the sequence is a GP with a common ratio \(r = \frac{2}{3}\). The first term might be a typo or a distractor.
To find the next term (\(t_5\)), we multiply the fourth term by the common ratio:
\(t_5 = t_4 \times r = \frac{1}{18} \times \frac{2}{3} = \frac{2}{54} = \frac{1}{27}\).
The correct option is (b) \(\frac{1}{27}\).
PART II: Short Answer Questions
8) If \(B \times A = \{(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)\}\) find A and B.
In the Cartesian product \(B \times A\), the set of all first elements in the ordered pairs is the set B, and the set of all second elements is the set A.
Given \(B \times A = \{(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)\}\).
The first elements are -2, 0, 3. So, \(B = \{-2, 0, 3\}\).
The second elements are 3, 4. So, \(A = \{3, 4\}\).
9) If \(A = \{-2, -1, 0, 1, 2\}\) and \(f : A \to B\) is an onto function defined by \(f(x) = x^2 + x + 1\) then find B.
Since \(f\) is an onto function, the range of \(f\) is equal to the codomain B. We need to find the image of each element in A.
The function is \(f(x) = x^2 + x + 1\).
- \(f(-2) = (-2)^2 + (-2) + 1 = 4 - 2 + 1 = 3\)
- \(f(-1) = (-1)^2 + (-1) + 1 = 1 - 1 + 1 = 1\)
- \(f(0) = (0)^2 + 0 + 1 = 1\)
- \(f(1) = (1)^2 + 1 + 1 = 3\)
- \(f(2) = (2)^2 + 2 + 1 = 4 + 2 + 1 = 7\)
The set of all images (the range) is \(\{1, 3, 7\}\).
Since \(f\) is onto, \(B = \text{Range of } f\).
Therefore, \(B = \{1, 3, 7\}\).
10) Compute x, such that \(10^4 \equiv x \pmod{19}\).
We need to find the remainder when \(10^4\) is divided by 19.
Let's compute powers of 10 modulo 19.
\(10^1 \equiv 10 \pmod{19}\)
\(10^2 = 100\). Dividing 100 by 19: \(100 = 5 \times 19 + 5\). So, \(10^2 \equiv 5 \pmod{19}\).
Now we can find \(10^4\):
\(10^4 = (10^2)^2 \equiv 5^2 \pmod{19}\)
\(10^4 \equiv 25 \pmod{19}\)
Now we find the remainder of 25 when divided by 19: \(25 = 1 \times 19 + 6\).
So, \(25 \equiv 6 \pmod{19}\).
Therefore, \(10^4 \equiv 6 \pmod{19}\).
\(x = 6\).
11) Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.
Let the Arithmetic Progression (A.P.) be \(a_1, a_2, a_3, a_4, a_5\).
We are given \(x, 10, y, 24, z\). So, \(a_1=x, a_2=10, a_3=y, a_4=24, a_5=z\).
In an A.P., the common difference \(d\) is constant. We can find \(d\) using \(a_2\) and \(a_4\).
\(a_4 = a_2 + (4-2)d \implies a_4 = a_2 + 2d\)
\(24 = 10 + 2d\)
\(14 = 2d \implies d = 7\)
Now we can find x, y, and z.
- \(x = a_1 = a_2 - d = 10 - 7 = 3\)
- \(y = a_3 = a_2 + d = 10 + 7 = 17\)
- \(z = a_5 = a_4 + d = 24 + 7 = 31\)
Thus, \(x=3, y=17, z=31\).
12) If \(1+2+3+\dots+K = 325\), then find \(1^3+2^3+3^3+\dots+K^3\).
We are given the sum of the first K natural numbers:
\(\sum_{n=1}^{K} n = 1+2+3+\dots+K = 325\)
The formula for this sum is \(\frac{K(K+1)}{2}\). So, \(\frac{K(K+1)}{2} = 325\).
We need to find the sum of the cubes of the first K natural numbers:
\(\sum_{n=1}^{K} n^3 = 1^3+2^3+3^3+\dots+K^3\)
The formula for the sum of cubes is \(\left(\frac{K(K+1)}{2}\right)^2\).
We can substitute the given value:
\(1^3+2^3+3^3+\dots+K^3 = (325)^2\)
\(325^2 = 105625\)
The value is 105625.
13) A Relation R is given by the set \(\{(x, y) / y = x+3, x \in \{0, 1, 2, 3, 4, 5\}\}\). Determine its domain and range.
The relation R is defined by the rule \(y = x+3\) where the input values for \(x\) are from the set \(\{0, 1, 2, 3, 4, 5\}\).
The domain of a relation is the set of all possible input values (x-values).
From the definition, the domain is explicitly given as Domain = \(\{0, 1, 2, 3, 4, 5\}\).
The range of a relation is the set of all resulting output values (y-values). We calculate \(y\) for each \(x\) in the domain:
- If \(x=0\), \(y = 0+3 = 3\)
- If \(x=1\), \(y = 1+3 = 4\)
- If \(x=2\), \(y = 2+3 = 5\)
- If \(x=3\), \(y = 3+3 = 6\)
- If \(x=4\), \(y = 4+3 = 7\)
- If \(x=5\), \(y = 5+3 = 8\)
The set of y-values is the range.
Range = \(\{3, 4, 5, 6, 7, 8\}\).
14) (Compulsory) Find the sum \(3 + 1 + \frac{1}{3} + \dots\).
The given series is \(3, 1, \frac{1}{3}, \dots\).
This is a Geometric Progression (GP). Let's find the first term \(a\) and the common ratio \(r\).
First term, \(a = 3\).
Common ratio, \(r = \frac{t_2}{t_1} = \frac{1}{3}\). (Check: \(t_3/t_2 = (1/3)/1 = 1/3\)).
Since the absolute value of the common ratio \(|r| = |\frac{1}{3}| < 1\), the sum to infinity of this GP exists.
The formula for the sum to infinity of a GP is \(S_\infty = \frac{a}{1-r}\).
Substituting the values of \(a\) and \(r\):
\(S_\infty = \frac{3}{1 - \frac{1}{3}} = \frac{3}{\frac{3-1}{3}} = \frac{3}{\frac{2}{3}}\)
\(S_\infty = 3 \times \frac{3}{2} = \frac{9}{2}\)
The sum of the series is \(\frac{9}{2}\) or 4.5.
PART III: Long Answer Questions
15) Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that \((A \cap B) \times C = (A \times C) \cap (B \times C)\).
First, let's list the elements of the sets A, B, and C.
- A = Set of natural numbers less than 8 = \(\{1, 2, 3, 4, 5, 6, 7\}\)
- B = Set of prime numbers less than 8 = \(\{2, 3, 5, 7\}\)
- C = Set of even prime number = \(\{2\}\)
Now, we will evaluate the Left-Hand Side (LHS): \((A \cap B) \times C\).
Step 1: Find \(A \cap B\).
\(A \cap B = \{1, 2, 3, 4, 5, 6, 7\} \cap \{2, 3, 5, 7\} = \{2, 3, 5, 7\}\)
Step 2: Find \((A \cap B) \times C\).
LHS = \(\{2, 3, 5, 7\} \times \{2\} = \{(2, 2), (3, 2), (5, 2), (7, 2)\}\)
Next, we will evaluate the Right-Hand Side (RHS): \((A \times C) \cap (B \times C)\).
Step 1: Find \(A \times C\).
\(A \times C = \{1, 2, 3, 4, 5, 6, 7\} \times \{2\} = \{(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)\}\)
Step 2: Find \(B \times C\).
\(B \times C = \{2, 3, 5, 7\} \times \{2\} = \{(2, 2), (3, 2), (5, 2), (7, 2)\}\)
Step 3: Find the intersection of the two sets.
RHS = \((A \times C) \cap (B \times C) = \{(2, 2), (3, 2), (5, 2), (7, 2)\}\)
Comparing the results:
LHS = \(\{(2, 2), (3, 2), (5, 2), (7, 2)\}\)
RHS = \(\{(2, 2), (3, 2), (5, 2), (7, 2)\}\)
Since LHS = RHS, the property is verified.
16) Let \(A = \{1, 2, 3, 4\}\) and \(B = \{2, 5, 8, 11, 14\}\) be two sets. Let \(f : A \to B\) be a function given by \(f(x) = 3x - 1\). Represent this function (i) by arrow diagram (ii) in a table form (iii) as a set of ordered pairs (iv) in a graphical form.
First, let's find the images of the elements of A under the function \(f(x) = 3x - 1\).
- \(f(1) = 3(1) - 1 = 2\)
- \(f(2) = 3(2) - 1 = 5\)
- \(f(3) = 3(3) - 1 = 8\)
- \(f(4) = 3(4) - 1 = 11\)
All images \(\{2, 5, 8, 11\}\) are in the codomain B.
(i) By arrow diagram:
An arrow diagram consists of two ovals representing the sets A and B, with arrows connecting each element of A to its corresponding image in B.
(Set A) (Set B)
1 \(\to\) 2
2 \(\to\) 5
3 \(\to\) 8
4 \(\to\) 11
14
(ii) In a table form:
A table with two rows (or columns) for x and f(x).
| x | f(x) = 3x - 1 |
|---|---|
| 1 | 2 |
| 2 | 5 |
| 3 | 8 |
| 4 | 11 |
(iii) As a set of ordered pairs:
The function is represented as a set of pairs (x, f(x)).
\(f = \{(1, 2), (2, 5), (3, 8), (4, 11)\}\)
(iv) In a graphical form:
We plot the ordered pairs as points on a Cartesian coordinate plane.
The points to be plotted are (1, 2), (2, 5), (3, 8), and (4, 11). This would be represented by four distinct points on a graph with an x-axis and a y-axis.
17) Find the sum of all natural numbers between 600 and 800 which are divisible by 11.
We need to find the sum of an arithmetic progression where the terms are multiples of 11 between 600 and 800.
Step 1: Find the first term (a).
Divide 600 by 11: \(600 \div 11 \approx 54.54\). The next integer is 55. So, the first number greater than 600 divisible by 11 is \(55 \times 11 = 605\). Thus, \(a = 605\).
Step 2: Find the last term (l).
Divide 800 by 11: \(800 \div 11 \approx 72.72\). The integer part is 72. So, the last number less than 800 divisible by 11 is \(72 \times 11 = 792\). Thus, \(l = 792\).
Step 3: Find the number of terms (n).
The common difference is \(d = 11\). Using the formula \(l = a + (n-1)d\):
\(792 = 605 + (n-1)11\)
\(792 - 605 = (n-1)11\)
\(187 = (n-1)11\)
\(n-1 = \frac{187}{11} = 17\)
\(n = 17 + 1 = 18\)
Step 4: Find the sum (S_n).
Using the formula \(S_n = \frac{n}{2}(a+l)\):
\(S_{18} = \frac{18}{2}(605 + 792)\)
\(S_{18} = 9(1397)\)
\(S_{18} = 12573\)
The sum of all natural numbers between 600 and 800 divisible by 11 is 12573.
18) Find the sum to n terms of the series \(5+55+555+\dots\).
Let \(S_n = 5 + 55 + 555 + \dots + n\) terms.
Step 1: Factor out the common term.
\(S_n = 5(1 + 11 + 111 + \dots + n \text{ terms})\)
Step 2: Multiply and divide by 9.
\(S_n = \frac{5}{9}(9 + 99 + 999 + \dots + n \text{ terms})\)
Step 3: Express each term as a power of 10 minus 1.
\(S_n = \frac{5}{9}((10-1) + (100-1) + (1000-1) + \dots + (10^n-1))\)
Step 4: Separate the terms into two groups.
\(S_n = \frac{5}{9}[(10 + 10^2 + 10^3 + \dots + 10^n) - (1 + 1 + 1 + \dots + n \text{ times})]\)
Step 5: Calculate the sum of each group.
The first group is a GP with \(a=10, r=10\). The sum is \(S_{GP} = \frac{a(r^n-1)}{r-1} = \frac{10(10^n-1)}{10-1} = \frac{10}{9}(10^n-1)\).
The second group is the sum of 'n' ones, which is simply \(n\).
Step 6: Substitute back into the expression for S_n.
\(S_n = \frac{5}{9}\left[ \frac{10}{9}(10^n-1) - n \right]\)
Simplifying the expression:
\(S_n = \frac{5 \times 10}{9 \times 9}(10^n-1) - \frac{5}{9}n\)
\(S_n = \frac{50}{81}(10^n-1) - \frac{5n}{9}\)
19) Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, ..., 24 cm. How much area can be decorated with these colour papers?
The sizes of the square papers are the side lengths. The area of a square with side 's' is \(s^2\).
The total area is the sum of the areas of all 15 squares:
Total Area = \(10^2 + 11^2 + 12^2 + \dots + 24^2\)
To find this sum, we can use the formula for the sum of the first 'n' squares: \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\).
We can write our sum as:
\((\sum_{k=1}^{24} k^2) - (\sum_{k=1}^{9} k^2)\)
Step 1: Calculate the sum of squares up to 24.
\(\sum_{k=1}^{24} k^2 = \frac{24(24+1)(2 \times 24+1)}{6} = \frac{24(25)(49)}{6}\)
\(= 4 \times 25 \times 49 = 100 \times 49 = 4900\)
Step 2: Calculate the sum of squares up to 9.
\(\sum_{k=1}^{9} k^2 = \frac{9(9+1)(2 \times 9+1)}{6} = \frac{9(10)(19)}{6}\)
\(= \frac{1710}{6} = 285\)
Step 3: Find the total area.
Total Area = \(4900 - 285 = 4615\)
The total area that can be decorated is 4615 cm².
20) The sum of three consecutive terms that are in A.P is 27 and their product is 288. Find the three terms.
Let the three consecutive terms in an A.P. be \(a-d, a, a+d\), where 'a' is the middle term and 'd' is the common difference.
Step 1: Use the sum to find 'a'.
Given, sum = 27.
\((a-d) + a + (a+d) = 27\)
\(3a = 27 \implies a = 9\)
So, the middle term is 9.
Step 2: Use the product to find 'd'.
Given, product = 288.
\((a-d)(a)(a+d) = 288\)
Substitute \(a=9\):
\((9-d)(9)(9+d) = 288\)
\((9-d)(9+d) = \frac{288}{9} = 32\)
Using the identity \((x-y)(x+y) = x^2 - y^2\):
\(9^2 - d^2 = 32\)
\(81 - d^2 = 32\)
\(d^2 = 81 - 32 = 49\)
\(d = \pm\sqrt{49} \implies d = \pm 7\)
Step 3: Find the three terms.
Case 1: If \(d=7\), the terms are \(a-d, a, a+d\) \(\Rightarrow\) \(9-7, 9, 9+7\) \(\Rightarrow\) \(2, 9, 16\).
Case 2: If \(d=-7\), the terms are \(a-d, a, a+d\) \(\Rightarrow\) \(9-(-7), 9, 9-7\) \(\Rightarrow\) \(16, 9, 2\).
In both cases, the set of terms is the same.
The three terms are 2, 9, and 16.
21) (Compulsory) If \(f(x) = x-4\), \(g(x) = x^2\) and \(h(x) = 3x-5\), prove that \((fog)oh = fo(goh)\).
We need to prove the associative property of function composition for the given functions.
LHS: \((fog)oh\)
First, find \(fog(x)\):
\(fog(x) = f(g(x)) = f(x^2) = x^2 - 4\)
Now, find \((fog)oh(x)\):
\((fog)oh(x) = (fog)(h(x)) = (fog)(3x-5)\)
Substitute \((3x-5)\) for \(x\) in the expression for \(fog(x)\):
\(= (3x-5)^2 - 4\)
\(= (9x^2 - 30x + 25) - 4\)
LHS = \(9x^2 - 30x + 21\)
RHS: \(fo(goh)\)
First, find \(goh(x)\):
\(goh(x) = g(h(x)) = g(3x-5) = (3x-5)^2\)
Now, find \(fo(goh)(x)\):
\(fo(goh)(x) = f(goh(x)) = f((3x-5)^2)\)
Substitute \((3x-5)^2\) for \(x\) in the expression for \(f(x)\):
\(= (3x-5)^2 - 4\)
\(= (9x^2 - 30x + 25) - 4\)
RHS = \(9x^2 - 30x + 21\)
Since LHS = RHS, we have proved that \((fog)oh = fo(goh)\).
PART IV: Construction
22) Construct a triangle similar to a given triangle PQR with its sides equal to \(\frac{7}{4}\) of the corresponding sides of the triangle PQR. (scale factor \(\frac{7}{4} > 1\))
Since the scale factor is \(\frac{7}{4}\), which is greater than 1, the new triangle (P'QR') will be larger than the original triangle (PQR).
Steps of Construction:
- Draw any triangle PQR (since no specific dimensions are given).
- Draw a ray QX starting from Q and making an acute angle with the side QR, on the side opposite to vertex P.
- On the ray QX, mark 7 (the greater of 7 and 4) equally spaced points. Label them \(Q_1, Q_2, Q_3, Q_4, Q_5, Q_6, Q_7\).
- Join \(Q_4\) (the point corresponding to the denominator of the scale factor) to the vertex R.
- Now, draw a line through \(Q_7\) (the point corresponding to the numerator) that is parallel to the line \(Q_4R\). This new line will intersect the extended side QR at a new point, R'. (To draw the parallel line, you can copy the angle at \(Q_4\) to \(Q_7\)).
- Next, draw a line through the new point R' that is parallel to the side PR. This line will intersect the extended side QP at a new point, P'.
- The triangle \(\triangle P'QR'\) is the required triangle, similar to \(\triangle PQR\), with sides that are \(\frac{7}{4}\) times the corresponding sides of \(\triangle PQR\).
Construct a triangle similar to a given triangle LMN with its sides equal to \(\frac{4}{5}\) of the corresponding sides of the triangle LMN. (scale factor \(\frac{4}{5} < 1\))
Since the scale factor is \(\frac{4}{5}\), which is less than 1, the new triangle (L'MN') will be smaller than the original triangle (LMN).
Steps of Construction:
- Draw any triangle LMN.
- Draw a ray MX starting from M and making an acute angle with the side MN, on the side opposite to vertex L.
- On the ray MX, mark 5 (the greater of 4 and 5) equally spaced points. Label them \(M_1, M_2, M_3, M_4, M_5\).
- Join \(M_5\) (the point corresponding to the denominator of the scale factor) to the vertex N.
- Now, draw a line through \(M_4\) (the point corresponding to the numerator) that is parallel to the line \(M_5N\). This new line will intersect the side MN at a new point, N'.
- Next, draw a line through the new point N' that is parallel to the side LN. This line will intersect the side LM at a new point, L'.
- The triangle \(\triangle L'MN'\) is the required triangle, similar to \(\triangle LMN\), with sides that are \(\frac{4}{5}\) times the corresponding sides of \(\triangle LMN\).